Question

A 20-inch wire will be cut into two parts. One part is bent to form a square and the other
into a circle. If the wire is cut x units from one end, express the sum of the areas of the figures formed
in terms of x.

Answer 1

A wire of 20 inches is cut x inches from one end

First part forms square of perimeter x inches

The other part forms circle of circumference (20 - x) inches as shown in the figure below

For the Square

$P = 4L$

$x = 4L$

$L = \frac{1}{4}x$

$Therefore, A = L ^2$

$A = (\frac{1}{4}x)^2$

Area of square in terms of x $= \frac{1}{16}x^2$

For the Circle

$Circumference, C = 2\pi r$

$20 - x = 2\pi r$

$r = \frac{20 - x}{2\pi}$

$Area, A = \pi r^2$

$= \frac{22}{7} × (\frac{20 - x}{2\pi} ) ^2$

$= \frac{22}{7} × \frac{(20 - x)^2× 49 }{4 × 484}$

$=\frac{1078 (20 - x)^2 }{13552}$

$=\frac{1078 (400 - 40x + x^2) }{13552}$

$=\frac{7 (400 - 40x + x^2) }{88}$

$= \frac{2800 - 280x + 7x^2}{88}$

Area of circle in terms of x $= \frac{350}{11} - \frac{35}{11} x + \frac{7}{88}x^2$

Sum of area of the figures formed $=\frac{1}{16}x^2 + \frac{7}{88}x^2 - \frac{35}{11} x +\frac{350}{11}$

$= (\frac{25}{176}x^2 - \frac{35}{11} x +\frac{350}{11} ) Sq. Inches$